[Swift]LeetCode343. 整数拆分 | Integer Break

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Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Note: You may assume that n is not less than 2 and not larger than 58.

---

给定一个正整数 n，将其拆分为至少两个正整数的和，并使这些整数的乘积最大化。 返回你可以获得的最大乘积。

示例 1:

输入: 2
输出: 1
解释: 2 = 1 + 1, 1 × 1 = 1。

示例 2:

输入: 10
输出: 36
解释: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36。

说明: 你可以假设 n 不小于 2 且不大于 58。

---

8ms

class Solution {
    func integerBreak(_ n: Int) -> Int {
        if n < 2 {
            return 0
        }
        
        var res = Array(repeating: 1, count: n+1)
        
        res[1] = 0
        
        for i in 2...n {
            var maxRes = 1
            for j in 1..<i {
                maxRes = max(maxRes, max(res[j], j) * max(i-j, res[i-j]))
            }
            res[i] = maxRes
        }
        return res[n]
    }
}

---

8ms

class Solution {
    func integerBreak(_ n: Int) -> Int {
        
        guard n > 3 else {
            return [1,1,1,2][n]
        }
        
        var times3 = n / 3
        
        if n % 3 == 1 {
            times3 -= 1
        }
        
        let times2 = (n - times3 * 3) / 2
        
        return Int(pow(3.0, Double(times3))) * Int(pow(2.0, Double(times2)))
    }
}

---

16ms

class Solution {
    func integerBreak(_ n: Int) -> Int {
        if n == 2 {
            return 1
        } else if n == 3 {
            return 2
        } else if n % 3 == 0 {
            return Int(pow(3, Double(n / 3)))
        } else if n % 3 == 1 {
            return Int(2 * 2 * pow(3, Double((n - 4) / 3)))
        } else { // 2
            return Int(2 * pow(3, Double((n - 2) / 3)))
        }
    }
}

---

24ms

class Solution {
    func integerBreak(_ n: Int) -> Int {
        var dps = Array(repeating: 0, count: n + 1)
        dps[1] = 1
        for num in 2...n {
            for j in 1..<num {
                dps[num] = max(dps[num], j * max(num - j, dps[num - j]))
            }
        }

        return dps[n]
    }
}